Question

A person travelling at 43.2km/h appplies the brake giving a deceleration of 6.0m/s2 to his scooter. How far will he travel before stopping?. Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Solution :

Initial velocity=u=43.2km/hr =43.2 x 5/18 =12 m/s
final velocity =V=0m/s [ As brakes are applied]
Acceleration=a= -6.0m/s2
Distance =s=?

Formula to be used : v²-u²=2as

s= v²-u²/2a
s=0²-12²/2x-6
S=12m

Hence distance travelled before stopping is 12m

Answer 2

Answer:

Initial velocity=u=43.2km/hr =43.2 x 5/18 =12 m/s

final velocity =V=0m/s [ As brakes are applied]

Acceleration=a= -6.0m/s2

Distance =s=?

Formula to be used : v²-u²=2as

⇒s= v²-u²/2a

⇒s=0²-12²/2x-6

⇒S=12m

∴Hence distance travelled before stopping is 12m